What is the average value of $3x^2+4x$ on the interval $2\leq x \leq 6$ ?
In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=3x^2+4x}$, ${a=2}$ and ${b=6}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{2}}^{ 6} ({3x^2+4x})\,dx}{ 6-{2}} \\\\ &=\dfrac{\Big[x^3+2x^2\Big]_{2}^6}{4} \\\\ &=\dfrac{288-16}{4} \\\\ &=68 \end{aligned}$ In conclusion, the average value of $3x^2+4x$ on the interval $2\leq x \leq 6$ is $68$.